Question

y''-2xy=0
Solve by using the ansatz:

Answer 1

\[y(x)=\sum_{n=0}^{\infty?}a _{n}x^n\]

Answer 2

Oh hell the original equation is y'-2xy=0

Answer 3

I get it until writing y'(x)=\[\sum_{n=1}^{\infty}na _{n}x^(n-1)\]

Answer 4

The other term will be
-2xy(x)=\[2\sum_{n=0}^{\infty}a _{n}x^(n+1)\]

Answer 5

now I guess we need to modify the summation to get the same powers of x, so that we can add them

Answer 6

I'm not sure what you're asking... can't \(\dot y -xy=0\) be easily solved usig separation of variables?

Answer 7

The object is to practice using power series. It is easily soluble by separation of variables and in this case, gives one an easy way to verify the final solution.

It's a nice question.

Answer 8

James is right, I can solve it with other technique easily, but I need to practice the power series with this. Can you help?

Answer 9

So if \( y = \sum_0^\infty a_n x^n \), then
\[ \begin{align}
y' &= \sum_0^\infty n a_n x^{n-1} &= a_1 + 2a_2x + 3a_3x^2 + 4a_4x^3 + ... \\
-2xy &= \sum_0^\infty -2a_nx^{n+1} &= 0 -2a_0x - 2a_1x^2 - 2a_2x^3 + ... \\
\end{align} \]
Hence
\[ 0 = y' - 2xy = a_1 + (2a_2 - 2a_0)x + (3a_3 - 2a_1)x^2 + (4a_4 - 2a_2)x^3 + ...\]
As the power series is identically zero, every coefficient is also zero, hence
\[ \begin{align}
0 &= a_1 \\
0 &= 2a_2 - 2a_0 \\
0 &= 3a_3 - 2a_1 \\
0 &= 4a_4 - 2a_2 \\
0 &= 5a_5 - 2a_3 \\
0 &= 6a_6 - 2a_4 \\
... \\
0 &= ja_j - 2a_{j-2} \\
\end{align} \]

Now as \( a_1 = 0 \) you can see that propagates forward all all odd index \( a_j = 0 \), i.e., \( 0 = a_1 = a_3 = a_5 = ... \).

Choose \( a_0 = 1 \). We can do that for now because we know that the original equation is linear and hence the scalar multiple of any solution is a solution. Choosing a value for \( a_0 \) just makes it more convenient to find one solution; we can multiply that solution by a scalar later for the general solution.

Then \(a_2 = 1 \),
\[ 4a_4 - 2a_2 = 0 \ \ \implies a_4 = 1/2 = 1/2!\]
\[ 6a_6 - 2a_4 = 0 \ \ \implies a_6 = 1/6 = 1/3!\]
and in general \( a_{2j} = 1/j! \)

Answer 10

Therefore
\[ y_c = \sum_{n=0}^\infty a_n x^n = \sum_{j=0}^\infty \frac{1}{j!} x^{2j} = e^{x^2} \]
is a solution of the equation and any function \( Cy_c \) for a constant C is also a solution.

Answer 11

Like it?

Answer 12

Yes very much. It is half algebra half logic, best part of maths!
Thanks

Answer 13

Oh sorry...I wasn't familiar with the term ansatz.